Problem

Find the area of the region enclosed between the …

02:52

Problem 48 Hard Difficulty

Let $A(n)$ be the area in the first quadrant enclosed by
the curves $y=\sqrt[n]{x}$ and $y=x$
(a) By considering how the graph of $y=\sqrt[n]{x}$ changes
as $n$ increases, make a conjecture about the limit of
$A(n)$ as $n \rightarrow+\infty$.
(b) Confirm your conjecture by calculating the limit.

Answer

a. $$
=\frac{2 n-(n+1)}{2(n+1)}=\frac{n-1}{2(n+1)}=\frac{1}{2}-\frac{1}{(n+1)}
$$
b. $$
\lim _{n \rightarrow \infty} A=1 / 2
$$

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Chapter 6

APPLICATIONS OF THE DEFINITE INTEGRAL IN GEOMETRY, SCIENCE, AND ENGINEERING

Section 1

Area Between Two Curves

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Video Transcript

Uh, probably the best way to do this balance to practice a few things. Now, Um, since we're only looking at the first quadrant, you can ignore everything to the right. I think it's important to know that the end through of X and the equation, like ALS X, will always intersect at one, because the end through it of one is always equal toe one. Um, so from 0 to 1 and the, um um, how would rewrite it as X to the one and power, Um, minus X dx. Um, so the other thing I would assume that X is gonna be fair, like, fairly large, because it's kind of the premise that's going on here, and, ah, as n gets bigger and just change colors, um, it gets to be like a wider curve, a wider curve, and then shoots over and a wider curve. Now it won't ever actually equal whatever value. But what I'm thinking about is how the area here for y equals X is always one half, which actually makes sense of you. You knew one half X to the second power from 0 to 1, and assuming you didn't do this piece um, which I guess would make sense because, uh, um if you plug in one one squared is one half of that. So the area highlighted right here is equal to one half, and my visual is trying to get at that. You're going to get really close to this answer being one exactly minus one half. Um, so the answer, as my conjecture is about to be, is one half as the limit ghost as Angus Infinity. Um, which, if we made this and value be ridiculously large. I don't know, Like I don't know. It's 100 sound large enough for you. Um, So what would happen is you would add one, 211 hundreds. What should be X to the 100 of one hundreds? Um, and then you have to multiply that are divided by the reciprocal or sorry divide by that number, which is multiplying by the reciprocal, which would be 100 of one over Ah 100. And you could see that this number is getting closer and closer one half, Um, because this is really, really close to one, which was my conjecture there, Um, minus one half is really, really close. to one half, so I'm going to stop right there. I think I've kind of over explained this. Um, so hopefully have said something that has either picture interest to go your own route or that I have explained this sufficiently.

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